Problem 73 - College Algebra and Calculus: An Applied Approach

Question

The distance between \(x\) and 5 is no more than \(3 .\)The distance between \(x\) and \(-10\) is at least \(6 .\)

Step-by-Step Solution

Answer

The values of \(x\) that satisfy both conditions are within the interval \([-4, 2]\).

Step 1: Determine the first inequality

The exercise says the distance between \(x\) and 5 is no more than 3. In mathematical terms, this can be represented as the absolute value inequality \(|x - 5| \leq 3\). This can be further written as two separate inequalities: \(x - 5 \leq 3\) and \(-x + 5 \leq 3\). Solving these yield \(x \geq 2\) and \(x \leq 8\). Thus, the first condition narrows the possible range of \(x\) down to \([2, 8]\).

Step 2: Determine the second inequality

Next, the exercise tells us that the distance between \(x\) and -10 is at least 6. It can be translated into the absolute value inequality \(|x - (-10)| \geq 6\), equivalent to \(|x + 10| \geq 6\). This can be separated into two inequalities: \(x + 10 \geq 6\) and \(-x - 10 \geq 6\). Solving them gives us \(x \geq -4\) and \(x \leq -16\). Thus the second condition reduces the possible range of \(x\) to \((-\infty, -16] \cup [-4, +\infty)\).

Step 3: Find the intersection of the two ranges

Finally, to meet both conditions simultaneously, \(x\) should fall within the intersection of the ranges obtained in steps 1 and 2. Comparing the two ranges, it can be noticed that only the interval \([-4, 2]\) falls within both ranges. Therefore, this is the final solution range for the problem.

Key Concepts

Inequality Solving

Solving inequalities involves finding all the values of a variable that satisfy a given condition. In this exercise, we have two absolute value inequalities to work with.

• The first inequality \( |x - 5| \leq 3 \) represents how close or far \(x \) can be from 5, ensuring the maximum distance is 3 units. We break this into \( x - 5 \leq 3 \) and \( x - 5 \geq -3 \), leading us to the solution \( 2 \leq x \leq 8 \).
• The second inequality \( |x + 10| \geq 6 \) describes a situation where \(x \) must be at least 6 units from -10. Dividing it gives two inequalities: \( x + 10 \geq 6 \) and \( x + 10 \leq -6 \), resulting in \( x \geq -4 \) or \( x \leq -16 \).

Through these steps, we develop a system of inequalities, which helps to narrow down the possible values of \(x \). The process involves breaking down absolute values and solving linear inequalities typically requiring simple algebraic manipulations.

Range Intersection

Understanding how to find the intersection of ranges is essential in inequality problems. It builds on separately solved inequalities.

• From step one, the range \( [2, 8] \) is derived from the first inequality.
• The second range \( (-\infty, -16] \cup [-4, +\infty) \) comes from the second inequality.

To identify the intersection, we look for common \(x \) values within these ranges. Specifically, we need to see where these two conditions overlap.
By carefully overlaying these ranges, it's clear that \[ -4, 2 \] satisfies both. The intersection represents the set of all potential solutions that meet every condition in the problem. Visualizing ranges on a number line can simplify this process, making it more intuitive to detect overlaps.

Distance in Algebra

In algebra, the concept of distance can transform into numeric conditions we solve through inequalities. The basic principle is to express how far one number is from another using absolute values.
In the context of this exercise:

• The expression \( |x - 5| \) translates to the distance between \(x \) and 5 on a number line. An inequality like \( |x - 5| \leq 3 \) indicates \(x \) is at most 3 units away.
• Similarly, \( |x + 10| \) describes how far \(x \) is from -10, with \( |x + 10| \geq 6 \) implying at least a 6-unit distance.

This method allows algebraic representations of distance to be converted into solvable mathematical problems. By comparing these distance expressions, we create conditions that describe the relationship between a variable and fixed values. This provides a deeper understanding of physical distances and how they're captured in equations.